What is the value of v 5

p value

In this post, we'll explain everything you can think of on the subjectp value should know. After we briefly show you what you need the p value for and what it does, we will guide you step-by-step through the calculation and show you how you can easily do it yourself.

So if you're wondering how"What does a small p value mean?" go through your head, then you've come to the right place. In our video we also explain the whole thing visually!

p value simply explained

The p value is an important part of the hypothesis test. Its main role is to help reject the null hypothesis, which is done by comparing it to the level of significance. If the p value turns out to be smaller when comparing with the significance level, then you can reject the null hypothesis and instead the alternative hypothesis accept. However, if the p value exceeds the significance level, you have to keep the null hypothesis for the time being. The p value can be calculated in two main ways: using the probability function of the binomial distribution or generally using the z-transformation and the normal distribution table.

Calculate p value

Using a detailed calculation example, we will explain in this section how you can calculate the p value by hand. There are different approaches. In this article we will show you how you can calculate the p-value for binomially distributed variables once with a low number of cases using the probability function and once with a high number of cases using the test statistics.

p value example

To do this, we will use an example. At German universities, around 28% of students drop out before their bachelor's degree and therefore drop out without a degree. The small private university "Schmetterling" wants to do something about these frightening numbers and is starting a program that guarantees and offers students intensive individual study advice several times during the semester. After an initial start-up phase, the university wants to check the drop-out rate on the assumption that less than 28% are dropping out.

In the course of the investigation, the following hypotheses are made:

: Individual study advice minimizes the drop-out rate of students to less than 28%

: The number of dropouts is 28%

Since the private university “Schmetterling” has very few students, only a small sample of 25 newly started students can be considered. Of these 25 students, after the start of the “individual course counseling” measure, only 4 dropped out, which corresponds to a rate of 16% and is therefore clearly below the nationwide value of 28%. But be careful: is this value really due to the measure “individual student counseling” or is it just a coincidence? Exactly for this reason we now calculate the p value to determine whether we can accept the H1 or we have to keep the H0 for the time being.

p value probability function

With such a small sample size, the p value can be determined relatively easily using the probability function. As a reminder, here is the written probability function:

Using this function, we now calculate how likely it is that only 4 or fewer (i.e. 4, 3, 2, 1, or 0) students drop out of their studies prematurely. These probabilities are then added up and we get the p value. Should this value then the significance level fall below, then we can reject the null hypothesis.

So first we calculate the individual probabilities for all values .

Now you just add up all your results for the final p value:

p value significance

When comparing with the significance level, it quickly becomes clear: the calculated p value is 13.04% and thus well above the estimated significance level of 5%.

We cannot therefore reject the null hypothesis for the time being, as we cannot say with sufficient certainty that our measured result came about through the measure “individual study advice” instead of by chance.

p value test statistic

The p value can, however, be calculated in another way; for larger numbers of cases using the so-called test statistics or test variable T. We will extend our example further and remain in the same subject area.

The large state university "Elefant" sees great potential in the idea of ​​the private university "Schmetterling" and is also introducing the measure "individual student counseling" in order to reduce the drop-out rate among students. In the first year after the introduction of the measure, 245 of 1000 new students dropped out, which corresponds to a rate of 24.5% and is again well below the usual 28%. It should now be checked again, but this time with a higher number of cases, whether the measure was decisive for the reduction or whether it could have been coincidence.

The hypotheses have been adjusted slightly and are now as follows:

: Individual study advice minimizes the drop-out rate of students to less than 28%

: Individual study advice does not minimize the drop-out rate of students to less than 28%

For the calculation of the p-value in the case of high numbers of cases, you first need several pieces of information: the mean and the standard deviation of the sample. These can be easily calculated using the following formulas:

The data obtained in this way are now used in the formula by which the test statistics are calculated and then the test variable is obtained. In addition, you need the value for k for this formula. It stands for the number of observed cases and is therefore 245 in this example.

After inserting all values, you get a result of -2.47 for the test variable T of the test statistics.

p value table

But what happens now to the value of -2.47? What does this actually say? When calculating the test statistic, we essentially did nothing other than perform a z-transformation. This enables us to look up this value in a table for the standard normal distribution and to check the probability of a value occurring that is the same or less. We remember: this probability must be below the significance level of lie so that the null hypothesis can be rejected.

With a value of -2.47 you have to select -2.4 on the vertical column and 0.07 on the horizontal column in order to be able to read off the probability of -2.47 in the end. This is 0.0068.

p value interpretation

Since our p value is significantly smaller than our significance level, we can reject the null hypothesis and the accept. For this reason, the state university "Elefant" can claim after carrying out its investigation: the measure "individual study advice" minimizes the drop-out rate of students to below 28%.

If our calculated p value were higher than failed, we should have retained the null hypothesis for the time being. So we could not have said that our measure was successful, but should still have assumed that the dropout rate averages 28%.